Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 2x}{x - 1} = \dfrac{3}{x - 1}$
Explanation: Multiply both sides by $x - 1$ $ \dfrac{x^2 + 2x}{x - 1} (x - 1) = \dfrac{3}{x - 1} (x - 1)$ $ x^2 + 2x = 3$ Subtract $3$ from both sides: $ x^2 + 2x - (3) = 3 - (3)$ $ x^2 + 2x - 3 = 0$ Factor the expression: $ (x + 3)(x - 1) = 0$ Therefore $x = -3$ or $x = 1$ At $x = 1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 1$, it is an extraneous solution.